## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 700: 41

#### Answer

$(\sqrt 3,\frac{\pi}{4},0.955)$.

#### Work Step by Step

Since $x=1, \, y=1, \, z=1$ and $$\rho=\sqrt{x^2+y^2+z^2},\quad \theta=\tan^{-1}( y/x), \quad \phi=\cos^{-1}(z/\rho),$$ then we have $$\rho=\sqrt{1+1+1 }= \sqrt 3,$$ $$\theta=\tan^{-1}1=\frac{\pi}{4},$$ $$\phi=\cos^{-1}(1/\sqrt 3)=0.955.$$ Hence, the spherical coordinates are $(\sqrt 3,\frac{\pi}{4},0.955)$.

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